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Math Help - Prove some identities! (Round two)

  1. #1
    MHF Contributor Drexel28's Avatar
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    Prove some identities! (Round two)

    And you thought it was over?


    Problem: Compute


    \displaystyle \sum_{n=2}^{\infty}\left(\zeta(n)-1\right)

    ,


    \displaystyle \sum_{n=1}^{\infty}\left(\zeta(2n)-1\right)


    and,


    \displaystyle \sum_{n=2}^{\infty}\left(\zeta(2n-1)-1\right)
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    Cool ones!

    \displaystyle{<br />
\sum_{n=2}^\infty (\zeta(n)-1) = \sum_{n=2}^\infty\sum_{m=2}^\infty \frac{1}{m^n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{m^n}<br />
}


    \displaystyle{=<br />
\sum_{m=2}^\infty \frac{1}{m^2-m} = \sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m}\right) = 1<br />
}


    Likewise:


    \displaystyle{<br />
\sum_{n=1}^\infty (\zeta(2n)-1) = \sum_{m=2}^\infty\sum_{n=1}^\infty \frac{1}{m^{2n}} = \sum_{m=2}^\infty\frac{1}{m^2-1}<br />
}

    \displaystyle{<br />
= \frac{1}{2}\sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m+1}\right) = \frac{3}{4}<br />
}

    And the last one is the difference between the first and the second, that is:

    \displaystyle{<br />
\sum_{n=1}^\infty (\zeta(2n+1)-1) = \frac{1}{4}<br />
}
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    Cool ones!

    \displaystyle{<br />
\sum_{n=2}^\infty (\zeta(n)-1) = \sum_{n=2}^\infty\sum_{m=2}^\infty \frac{1}{m^n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{m^n}<br />
}


    \displaystyle{=<br />
\sum_{m=2}^\infty \frac{1}{m^2-m} = \sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m}\right) = 1<br />
}


    Likewise:


    \displaystyle{<br />
\sum_{n=1}^\infty (\zeta(2n)-1) = \sum_{m=2}^\infty\sum_{n=1}^\infty \frac{1}{m^{2n}} = \sum_{m=2}^\infty\frac{1}{m^2-1}<br />
}

    \displaystyle{<br />
= \frac{1}{2}\sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m+1}\right) = \frac{3}{4}<br />
}

    And the last one is the different between the first and the second, that is:

    \displaystyle{<br />
\sum_{n=1}^\infty (\zeta(2n+1)-1) = \frac{1}{4}<br />
}
    Right, not too hard, but interesting indeed. Of course using the same methodology shows that \displaystyle \sum_{n=1}^{\infty}(-1)^n\left(\zeta(n)-1\right) =\frac{1}{2}. This one is slightly harder. Try


    \displaystyle \sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n}
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    Quote Originally Posted by Drexel28 View Post
    Right, not too hard, but interesting indeed. Of course using the same methodology shows that \displaystyle \sum_{n=1}^{\infty}(-1)^n\left(\zeta(n)-1\right) =\frac{1}{2}. This one is slightly harder. Try


    \displaystyle \sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n}
    Of course, the anternating one is just the difference between the second and the last ones you proposed in the first post.

    I find it very interesting that they're rational.

    The other one you proposed is the same method:

    \displaystyle{<br />
\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{nm^n} = \sum_{m=2}^\infty\left(\ln m - \ln (m-1) - \frac{1}{m}\right)<br />
}


    \displaystyle{<br />
=1+\lim_{n\rightarrow\infty} \left(\ln(n)-\sum_{m=1}^n\frac{1}{m}\right) = 1-\gamma<br />
}
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    Of course, the anternating one is just the difference between the second and the last ones you proposed in the first post.

    I find it very interesting that they're rational.

    The other one you proposed is the same method:

    \displaystyle{<br />
\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{nm^n} = \sum_{m=2}^\infty\left(\ln m - \ln (m-1) - \frac{1}{m}\right)<br />
}


    \displaystyle{<br />
=1+\lim_{n\rightarrow\infty} \left(\ln(n)-\sum_{m=1}^n\frac{1}{m}\right) = 1-\gamma<br />
}
    Right. How about this one


    \displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(2n+1)}
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Right. How about this one


    \displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(2n+1)}
    Spoiler:

    We know  \displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} .

    Thus we see  \displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(2n+1)} = \int_0^1 \log(\pi x) dx - \int_0^1 \log(\sin\pi x) dx = \log\pi-1+\log2 = \log(2\pi)-1 .
    Last edited by chiph588@; December 30th 2010 at 03:12 PM.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    We know  \displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} .

    Thus we see  \displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(2n+1)} = \int_0^1 \log(\pi x) dx - \int_0^1 \log(\sin\pi x) dx = \log\pi-1+\log2 = \log(2\pi)-1 .

    Right! Now someone else post a problem so I can solve one! haha
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  8. #8
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    Here are some nice ones. Compute:


    \displaystyle<br />
\sum_{n=1}^\infty \frac{|\mu (n)|}{n^s}


    \displaystyle<br />
\sum_{n=1}^\infty \frac{\sigma(n)}{n^s}


    \displaystyle<br />
\sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s}


    where \mu(n) is the Mobius function, \sigma(n) is the divisor function, and \omega(n) counts the number of distinct primes dividing n.

    Edit: just to clarify, by \sigma(n) I was referring, as usually denoted by this letter, to the function returning the sum of the divisors of n, and not the number of divisors, which is also called the divisor function.
    Last edited by Unbeatable0; December 30th 2010 at 02:58 PM.
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Unbeatable0 View Post

    \displaystyle<br />
\sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s}
    Note that we require  \text{Re}(s) > 1 . I will do one, because they all can be derived in a similar fashion.

    Spoiler:

    It's not too hard to see  2^\omega(n) is multiplicative. Hence

     \begin{aligned} \displaystyle \sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s} &= \prod_p\sum_{k=0}^\infty \frac{2^\omega(p^k)}{p^{sk}}\\<br />
&= \prod_p\left(1+2\sum_{k=1}^\infty\frac1{p^{sk}}\ri  ght)\\<br />
&= \prod_p\left(1+2\frac{p^{-s}}{1-p^{-s}}\right)\\<br />
&= \prod_p \frac{1+p^{-s}}{1-p^{-s}}\\<br />
&= \prod_p \frac{1-p^{-2s}}{(1-p^{-s})^2}\\<br />
&= \frac{\zeta^2(s)}{\zeta(2s)} \end{aligned}


    Try this one: Compute
     \displaystyle \int_0^1 (-\log x)^a dx
    Last edited by chiph588@; December 30th 2010 at 04:31 PM.
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  10. #10
    Super Member Random Variable's Avatar
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    What's required to switch the order of summation? Absolute convergence of the inner sum?

    Quote Originally Posted by Unbeatable0 View Post
    Cool ones!

    \displaystyle{<br />
\sum_{n=2}^\infty (\zeta(n)-1) = \sum_{n=2}^\infty\sum_{m=2}^\infty \frac{1}{m^n} = \sum_{m=2}^\infty\sum_{n=2}^\infty \frac{1}{m^n}<br />
}


    \displaystyle{=<br />
\sum_{m=2}^\infty \frac{1}{m^2-m} = \sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m}\right) = 1<br />
}


    Likewise:


    \displaystyle{<br />
\sum_{n=1}^\infty (\zeta(2n)-1) = \sum_{m=2}^\infty\sum_{n=1}^\infty \frac{1}{m^{2n}} = \sum_{m=2}^\infty\frac{1}{m^2-1}<br />
}

    \displaystyle{<br />
= \frac{1}{2}\sum_{m=2}^\infty \left(\frac{1}{m-1}-\frac{1}{m+1}\right) = \frac{3}{4}<br />
}

    And the last one is the difference between the first and the second, that is:

    \displaystyle{<br />
\sum_{n=1}^\infty (\zeta(2n+1)-1) = \frac{1}{4}<br />
}
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  11. #11
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     \displaystyle  \sum_{n=1}^{\infty} \frac{\sigma(n) }{n^s }


    I think all of us would factorise the summation as this arithmetic function is multiplicative !

     \displaystyle  = \prod_{\text{all primes p }} \left( 1 + \frac{\sigma(p)}{p^s} + \frac{\sigma(p^2)}{p^{2s}} + ... \right)

      \displaystyle = \prod_{\text{all primes p }} \left[ \sum_{k=0}^{\infty} \left(\frac{ p^{k+1} - 1 }{p^{ks}(p - 1) } \right) \right]

     \displaystyle  =  \prod_{\text{all primes p }}  \frac{1}{p-1} \left( \frac{p}{1 - 1/p^{s-1} } - \frac{1}{1 - 1/p^s } \right)


      \displaystyle =  \prod_{\text{all primes p }}  \frac{1}{p-1} \left( \frac{p-1}{ (1 - 1/p^{s-1} )(1 - 1/p^s) } \right)


      \displaystyle =  \prod_{\text{all primes p }} \frac{1}{(1 - 1/p^{s-1} )(1 - 1/p^s) }

      \displaystyle = \zeta(s) \zeta(s-1)


    About the first problem , I remember it was first discovered by Ramanujan , he expanded the product :

      \displaystyle  \frac{1}{\zeta(s)} = \prod_{\text{all primes p }} \left( 1 - \frac{1}{p^s} \right)

    As each factor provides only a prime but not a power of it , all the summands of the expansion should be square-free , and the signs of them exactly indicate the moebius function .

    Therefore ,    \displaystyle  \frac{1}{\zeta(s) } = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}

    But this time , absolute value is taken :

       \displaystyle  \sum_{n=1}^{\infty} \frac{|\mu (n)|}{n^s} =  \prod_{\text{all primes p }}  \left( 1 + \frac{1}{p^s} \right) = \frac{ \zeta(s)}{\zeta(2s)}
    Last edited by simplependulum; December 30th 2010 at 06:10 PM.
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  12. #12
    Super Member Random Variable's Avatar
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     \int^{1}_{0} (-\ln x)^{a} \ dx

     = (-1)^{a} \int^{1}_{0} (\ln x)^{a} \ dx

    let  x = e^{-u}

     = (-1)^{a} \int_{\infty}^{0} (-u)^{a} (-e^{-u}) \ du

     = \int^{\infty}_{0} u^{a} e^{-u}   \ du

     = \Gamma (a+1)
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  13. #13
    MHF Contributor chiph588@'s Avatar
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    Evaluate  \displaystyle \int_0^1 \int_0^1 \frac{\log(xy)}{xy-1} dxdy .
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  14. #14
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Evaluate  \displaystyle \int_0^1 \int_0^1 \frac{\log(xy)}{xy-1} dxdy .
    Damnit, I had to go and all the cool number theory ones were already done!


    Spoiler:



    [CENTER]


    This particular case falls to pretty easy techniques

    \displaystyle \begin{aligned}\int_0^1\int_0^1\frac{\ln(xy)}{1-xy}\text{ }dx\text{ }dy &= \int_0^1\int_0^1 \sum_{n=0}^{\infty}\ln^s(xy)(xy)^n\text{ }dx\text{ }dy\\ &=\sum_{n=0}^{\infty}\int_0^1\int_0^1 \ln(xy)(xy)^n\text{ }dx\text{ }dy\\ &= \sum_{n=0}^{\infty}\int_0^1 \frac{y^n\left((n+1)\log(y)-1\right)}{(n+1)^2}\text{ }dy\\ &= -\sum_{n=0}^{\infty}\frac{2}{(n+1)^3}\\ &= -2\zeta(3)\end{aligned}


    But, it's slightly harder (not much, ask if you're interested) that


    \displaystyle \int_0^1\int_0^1\frac{\ln^s(xy)}{1-xy}\text{ }dx\text{ }dy=(-1)^s\Gamma(s+2)\zeta(s+2)\quad \text{Re}(s)>1


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    Spoiler:
     \displaystyle  \int_0^1 \int_0^1 \frac{\ln^s(xy)}{xy - 1  }~dx dy

    Sub.  xy = u (  y being constant ) we have

     \displaystyle  \int_0^1 \int_0^1 \frac{\ln^s(xy)}{xy - 1  }~dx~dy

     \displaystyle  = \int_0^1 \int_0^y \frac{\ln^s(u)}{u-1} \frac{du}{y} ~dy

     \displaystyle  = \int_0^1 \int_u^1 \frac{dy}{y}~ \frac{\ln^s(u)}{u-1}~du

     \displaystyle  = - \int_0^1 \frac{\ln^{s+1}(u)}{u-1} ~du

    Sub.  \displaystyle  u = e^{-t}

    \displaystyle   = - \int_0^{\infty} \frac{(-1)^{s+1} t^{s+1} }{ 1 - e^t }~dt

     \displaystyle  = (-1)^{s+1} \int_0^{\infty} \frac{t^{s+1}}{e^t - 1 } ~dt

     \displaystyle  = (-1)^{s+1} \Gamma(s+2) \zeta(s+2)



    How about these ?

     \displaystyle \int_0^1 \int_0^1 \int_0^1 \frac{dxdydz}{(1 + x^2 + y^2 + z^2)^2 } and

     \displaystyle  M(x,y) = \frac{\pi}{4} ~ \frac{x+y}{ \big K\left[ \left( \frac{x-y}{x+y} \right)^2 \right] }

    where  \displaystyle  K(m) = \int_0^{\pi/2} \frac{dx}{\sqrt{ 1 - m \sin^2(x) } } and  M(x,y) is arithmetic geometric mean :

      \displaystyle  a_1 = \frac{x+y}{2} ~,~ g_1 = \sqrt{xy} Then

    \displaystyle   a_{n+1} = \frac{a_n + g_n }{2} ~,~ g_{n+1} = \sqrt{a_n g_n }

     \displaystyle   M(x,y) = \lim_{n\to \infty} a_n =  \lim_{n\to \infty} g_n

    EDIT : I have made a very big mistake on the first problem , the integrand should be squared !
    Last edited by simplependulum; December 31st 2010 at 08:20 PM.
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